Problem: $\int^{\pi/4}_{0}\sin(4x)\,dx\, = $
Answer: Strategy Let's first find the indefinite integral $\int\sin(4x)\,dx\, $. Then we can use the result to compute the definite integral. Finding the indefinite integral To find $\int\sin(4x)\,dx\, $, we can use U-substitution. If we let $ {u=4x}$, then ${du=4 \, dx}$ and ${ \,dx=\dfrac{1}{4}\, du}$. So we have: $\begin{aligned}\int\sin({4x})\,{dx}\, &=\int \sin( u)\cdot {\dfrac14\, du}\,\\\\\\\\ &=\dfrac14\int\sin(u)\,du\\\\\\\\ &=\dfrac{1}{4}\cdot -\cos(u)+C\\\\\\\\ &=-\dfrac14\cos(4x)+C \end{aligned}$ Evaluating the definite integral Now let's compute the definite integral: ∫ π / 4 0 sin ( 4 x ) d x = − 1 4 cos ( 4 x ) ∣ ∣ ∣ π / 4 0 = − 1 4 ( cos ( 4 ⋅ π 4 ) − cos ( 4 ⋅ 0 ) ) = − 1 4 ( cos ( π ) − cos ( 0 ) ) = − 1 4 ( − 1 − 1 ) = 1 2 \begin{aligned}\int^{\pi/4}_{0}\sin(4x)\,dx\,&= -\dfrac14\cos(4x)\Bigg|^{\pi/4}_0\\\\\\\\ &=-\dfrac{1}{4}\left(\cos\left(4\cdot\dfrac{\pi}{4}\right)-\cos(4\cdot 0)\right)\\\\\\\\ &=-\dfrac14(\cos(\pi)-\cos(0))\\\\\\\ &=-\dfrac14(-1-1)\\\\\\\\ &=\dfrac12\end{aligned} [Did we have to find the indefinite integral first?] The answer $\int^{\pi/4}_{0}\sin(4x)\,dx\, = \dfrac12$